Step 1: Expression for Acceleration.

The force acting on the electron due to the electric field is given by \( F = eE \), and according to Newton's second law, the acceleration \( a \) can be expressed as: \[ a = \frac{F}{m} = \frac{eE}{m} \]

Step 2: Conclusion.

Thus, the acceleration of the electron is \( \frac{Ee}{m} \), which means the correct answer is (ii).
Quick Tip: The acceleration of a charged particle in an electric field is directly proportional to the charge and the field strength.

Step 1: Equivalent Resistance for Series and Parallel Configurations.

For three identical resistors \( R \), the total resistance when connected in series is: \[ R_1 = 3R \]
The total resistance when connected in parallel is: \[ R_2 = \frac{R}{3} \]

Step 2: Relating \( R_1 \) and \( R_2 \).

We are given that \( R_1 = nR_2 \). Substituting the expressions for \( R_1 \) and \( R_2 \), we get: \[ 3R = n \times \frac{R}{3} \]

Step 3: Solving for \( n \).

Simplifying the equation: \[ 3R = \frac{nR}{3} \quad \Rightarrow \quad n = 9 \]

Step 4: Conclusion.

Thus, the minimum value of \( n \) is \( 9 \), so the correct answer is (iv).
Quick Tip: For resistors in series, the total resistance is the sum of individual resistances, and for parallel resistors, the total resistance is the reciprocal of the sum of reciprocals.

Step 1: Formula for the Radius of the Circular Path.

The radius \( r \) of the circular trajectory of a charged particle in a magnetic field is given by: \[ r = \frac{mv}{qB} \]
where \( m \) represents the particle's mass, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

Step 2: Analyzing the Proportionality.

The momentum \( p \) of the particle can be written as \( p = mv \), so the radius becomes: \[ r = \frac{p}{qB} \]
This shows that the radius is directly proportional to the momentum \( p \) of the particle.

Step 3: Conclusion.

Thus, the correct answer is (ii) momentum of the particle.
Quick Tip: The radius of the circular path of a charged particle in a magnetic field depends on its momentum and the strength of the magnetic field.

Step 1: Understanding the AND Gate Behavior.

An AND gate gives an output of 1 only if all its inputs are 1. Let's examine each input combination:

- (i) A = 1, B = 0, C = 0: Output is 0 since at least one input is 0.

- (ii) A = 0, B = 1, C = 0: Output is 0 because A = 0.

- (iii) A = 1, B = 0, C = 1: Output is 1, assuming only A and C are connected to the gate.

- (iv) A = 1, B = 1, C = 0: Output is 0 because C = 0.


Step 2: Conclusion.

Therefore, the correct input combination that produces an output of 1 from the AND gate is (iii) A = 1, B = 0, C = 1.
Quick Tip: An AND gate outputs 1 only when all its inputs are 1. If any input is 0, the output will be 0.

Step 1: Nuclear Reaction Analysis.

The given nuclear reaction involves a nitrogen-7 nucleus interacting with a neutron. This results in the formation of a carbon-14 nucleus and an unknown particle \( X \).


Step 2: Balancing the Reaction.

To determine the identity of \( X \), we apply conservation of atomic number and mass number.
The nitrogen nucleus has atomic number 7 and mass number 14, while the neutron has atomic number 0 and mass number 1.
The carbon-14 nucleus formed has atomic number 6 and mass number 14.

To conserve both atomic and mass numbers, the missing particle \( X \) must have atomic number 1 and mass number 1 — which corresponds to a proton.

Step 3: Conclusion.

Therefore, the unknown particle \( X \) is a proton, so the correct answer is (i).
Quick Tip: In nuclear reactions, the sum of the atomic numbers and mass numbers on both sides must be equal. This is crucial for balancing the reactions.

Step 1: Electromagnetic Wave Propagation.

The given wave equation for the electric field is in the form \( E = E_0 \cos(\omega t + kx) \), where \( E_0 \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.


Step 2: Relating to Refractive Index.

The refractive index \( n \) of a medium is related to the speed of light in the medium and the speed of light in vacuum. It is given by: \[ n = \frac{c}{v} \]
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. The wave number \( k \) is related to the speed of light in the medium by: \[ k = \frac{\omega}{v} \]
By comparing the wave equation given, we find that the refractive index is \( n = 2.0 \).


Step 3: Conclusion.

Therefore, the refractive index is \( 2.0 \), making the correct answer (ii).
Quick Tip: The refractive index of a medium can be calculated using the wave number and angular frequency, and it indicates how much the wave is slowed in the medium compared to vacuum.

Step 1: Formula for Energy Stored in a Capacitor.

The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \]
where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference across the plates.


Step 2: Relationship Between Charge and Voltage.

The charge \( Q \) on the plates of the capacitor is related to the capacitance and the voltage by the equation: \[ Q = C V \]

Step 3: Substituting Voltage Expression.

From the energy formula \( U = \frac{1}{2} C V^2 \), we can solve for \( V \): \[ V = \sqrt{\frac{2U}{C}} \]

Step 4: Final Expression for Charge.

Substituting this value of \( V \) into the equation for charge: \[ Q = C \times \sqrt{\frac{2U}{C}} = \sqrt{2CU} \] Quick Tip: The charge on the plates of a capacitor is related to its capacitance and the energy stored in it through the expression \( Q = \sqrt{2CU} \).

Step 1: Definition of Photoelectric Effect.

The photoelectric effect refers to the phenomenon where electrons are ejected from a material (usually metal) when it is exposed to light or other electromagnetic radiation.


Step 2: Explanation of the Effect.

When photons of sufficient energy strike the surface of the material, they transfer their energy to the electrons in the material. If the energy of the photons is greater than the work function of the material (the minimum energy required to remove an electron from the surface), the electron is ejected from the material.


Step 3: Conclusion.

The photoelectric effect demonstrates the particle nature of light, where light behaves as particles (photons) that carry energy. Quick Tip: The photoelectric effect shows that light has particle-like properties and that the energy of the incident photons must exceed the work function of the material to release electrons.

Step 1: N-type Semiconductor.

In an n-type semiconductor, the material is doped with a group V element (such as phosphorus), which has five valence electrons. These extra electrons are free to move and contribute to electrical conduction.


Step 2: Majority Charge Carriers.

The majority charge carriers in an n-type semiconductor are electrons. These are the free electrons provided by the donor atoms.


Step 3: Minority Charge Carriers.

The minority charge carriers in an n-type semiconductor are holes. These are the absence of electrons in the valence band, which behave like positive charge carriers.


Step 4: Conclusion.

Thus, in an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. Quick Tip: In n-type semiconductors, electrons are the majority carriers, and holes are the minority carriers.

Step 1: Understanding Potentiometer.

A potentiometer is a device used to measure the potential difference by comparing it with a known reference voltage. The null deflection length is the length of the wire over which the potential gradient is balanced by the reference voltage.


Step 2: Effect of Decreasing Potential Gradient.

The potential gradient is defined as the potential difference per unit length of the wire. If the potential gradient is decreased, it means that for a given length of wire, the potential difference is smaller. To achieve null deflection, the length of the wire required will increase. Thus, the null deflection length increases when the potential gradient decreases.


Step 3: Conclusion.

Decreasing the potential gradient will increase the null deflection length. Quick Tip: In a potentiometer, decreasing the potential gradient results in a longer null deflection length, as the potential difference per unit length decreases.

Step 1: Galvanometer and Voltmeter.

A galvanometer is an instrument used to measure small currents, whereas a voltmeter is used to measure potential differences (voltages) across two points in a circuit.


Step 2: Conversion Process.

To convert a galvanometer into a voltmeter, a high-value resistor is connected in series with the galvanometer. The high resistance ensures that only a small current flows through the galvanometer when a potential difference is applied, allowing the voltmeter to measure larger voltages. The value of the series resistor is chosen based on the range of voltages to be measured.


Step 3: Conclusion.

Thus, a galvanometer is converted into a voltmeter by connecting a high-value resistance in series with it. Quick Tip: A high-value resistor in series with a galvanometer enables it to measure voltages by limiting the current passing through it.

Step 1: Formula for Minimum Deviation.

The angle of minimum deviation \( D_{min} \) in a prism is related to the prism angle \( A \) and the refractive index \( n \) by the following formula: \[ \sin \left( \frac{D_{min}}{2} \right) = \frac{n-1}{\sin \left( \frac{A}{2} \right)} \]

Step 2: Substituting Given Values.

Given the prism angle \( A = 10^\circ \) and refractive index \( n = 2 \), we can substitute these values into the formula: \[ \sin \left( \frac{D_{min}}{2} \right) = \frac{2-1}{\sin \left( \frac{10^\circ}{2} \right)} = \frac{1}{\sin 5^\circ} \]

Step 3: Solving for \( D_{min} \).

Using \( \sin 5^\circ \approx 0.0872 \), we get: \[ \sin \left( \frac{D_{min}}{2} \right) = \frac{1}{0.0872} \approx 11.46 \]

Since the sine function cannot exceed 1, this indicates that the given parameters result in an impractical situation, where the angle of minimum deviation cannot be achieved under normal conditions.


Step 4: Conclusion.

The given combination of prism angle and refractive index does not yield a physically valid solution for the angle of minimum deviation. Quick Tip: The angle of minimum deviation for a prism is constrained by the sine function, which cannot exceed 1. Ensure the values for angle and refractive index are practical.

Step 1: Intensity in Young's Double-Slit Experiment.

The intensity in Young's double-slit experiment is given by: \[ I = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) \]
where \( I_0 \) is the maximum intensity, \( d \) is the distance between the slits, \( \lambda \) is the wavelength of light, and \( \theta \) is the angle of the screen.

The maximum intensity occurs when the phase difference is an integer multiple of \( 2\pi \), and the minimum intensity occurs when the phase difference is an odd multiple of \( \pi \).

Step 2: Intensity Ratio.

The ratio of the maximum intensity (\( I_{max} \)) to the minimum intensity (\( I_{min} \)) is given as 9:1. Therefore, \[ \frac{I_{max}}{I_{min}} = 9 \]

Since the intensity ratio is related to the square of the cosine function of the phase difference, we can use the relation for the maximum and minimum intensities. The ratio of intensities also depends on the slit width, which affects the diffraction pattern and the fringe visibility. To satisfy the given intensity ratio, the slit width must be adjusted accordingly.

Step 3: Conclusion.

The required ratio of the slit width for the given intensity ratio will be determined by specific experimental conditions, which balance the diffraction and interference effects. The detailed calculation would require further information about the wavelength and distance between the slits. Quick Tip: In Young's double-slit experiment, the intensity ratio is influenced by the slit width and the fringe visibility, which are interdependent on the wavelength and the distance between slits.

Step 1: Resistance of a Wire.

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.

Step 2: Effect of Stretching the Wire.

When the wire is stretched, its length increases, and its radius decreases. If the radius is halved, the cross-sectional area \( A \), which is proportional to the square of the radius (\( A \propto r^2 \)), will decrease by a factor of 4.

The resistance is directly proportional to the length and inversely proportional to the area. If the length increases by a factor of \( k \) (since stretching the wire increases the length), and the area decreases by a factor of 4, the new resistance \( R_{new} \) can be calculated as: \[ R_{new} = R \times \left( \frac{L_{new}}{L} \right) \times \left( \frac{A}{A_{new}} \right) = 2 \times 4 = 8 \, \Omega \]

Step 3: Conclusion.

The new resistance of the wire is \( 8 \, \Omega \). Quick Tip: When a wire is stretched, its resistance increases due to the increase in length and the decrease in cross-sectional area.

Step 1: Formula for Period of Revolution.

The period \( T \) of revolution of a charged particle moving in a magnetic field is given by the formula: \[ T = \frac{2 \pi m}{q B} \]
where \( m \) is the mass of the particle, \( q \) is the charge of the particle, and \( B \) is the magnetic field strength.

Step 2: Mass and Charge of the Proton and \( \alpha \)-Particle.

- For the proton, the charge \( q_p = +e \) (where \( e \) is the elementary charge) and mass \( m_p \) is the mass of a proton.
- For the \( \alpha \)-particle, the charge \( q_{\alpha} = 2e \) (twice the charge of a proton) and mass \( m_{\alpha} = 4m_p \) (four times the mass of a proton).

Step 3: Ratio of Periods.

The ratio of the periods \( T_p \) for the proton and \( T_{\alpha} \) for the \( \alpha \)-particle is given by: \[ \frac{T_{\alpha}}{T_p} = \frac{m_{\alpha} / q_{\alpha}}{m_p / q_p} = \frac{4m_p / 2e}{m_p / e} = 2 \]

Step 4: Conclusion.

The ratio of the period of revolution of the \( \alpha \)-particle to the proton is 2:1. Quick Tip: The period of revolution depends on the mass and charge of the particle. A heavier and more charged particle will have a larger period of revolution.

Step 1: Stopping Potential and Intensity.

The stopping potential in the photoelectric effect depends on the energy of the incident photons, which in turn depends on the frequency of light. The stopping potential is given by: \[ V = \frac{h f}{e} - W \]
where \( h \) is Planck's constant, \( f \) is the frequency of the light, \( e \) is the charge of an electron, and \( W \) is the work function of the material.

Step 2: Effect of Distance on Intensity.

The intensity of light decreases with the square of the distance from the source. The intensity \( I \) is inversely proportional to the square of the distance \( r \), i.e., \[ I \propto \frac{1}{r^2} \]

Since the energy of individual photons is not affected by the distance, the stopping potential remains unchanged. The decrease in intensity (due to the increased distance) will result in fewer electrons being emitted, but the energy of the emitted electrons (and thus the stopping potential) remains the same.

Step 3: Conclusion.

The stopping potential remains the same even if the source is moved to \( 3r \) because the energy of the photons is not affected by the distance. Quick Tip: The stopping potential in the photoelectric effect depends on the energy of the incident photons, which is not affected by the distance from the light source.

Step 1: Formula for Potential Energy.

The total electric potential energy \( U \) of a system of charges is given by the formula: \[ U = \sum_{iwhere \( k \) is Coulomb's constant, \( q_i \) and \( q_j \) are the charges, and \( r_{ij} \) is the distance between the charges.

Step 2: Calculate Potential Energy for the Three Charges.

For the given system, the charges are \( +2q \), \( -q \), and \( -q \). The distance between any two charges is \( a \). Thus, the potential energy between each pair of charges is:
- Between \( +2q \) and \( -q \): \[ U_{12} = \frac{k \times 2q \times (-q)}{a} = -\frac{2kq^2}{a} \]
- Between \( -q \) and \( -q \): \[ U_{23} = \frac{k \times (-q) \times (-q)}{a} = \frac{kq^2}{a} \]
- Between \( +2q \) and \( -q \) (again): \[ U_{13} = \frac{k \times 2q \times (-q)}{a} = -\frac{2kq^2}{a} \]

Step 3: Total Potential Energy.

The total potential energy is the sum of all pairwise potential energies: \[ U_{total} = U_{12} + U_{23} + U_{13} = -\frac{2kq^2}{a} + \frac{kq^2}{a} - \frac{2kq^2}{a} = -\frac{3kq^2}{a} \]

Step 4: Conclusion.

Thus, the total electric potential energy of the system is \( U = -\frac{3kq^2}{a} \). Quick Tip: The potential energy of a system of charges depends on the pairwise interactions between all charges, and it is calculated using Coulomb's law.

Step 1: Magnetic Moment of a Coil.

The magnetic moment \( M \) of a single coil is given by: \[ M = i A \]
where \( i \) is the current, and \( A \) is the area of the coil. For a circular coil of radius \( r \), the area \( A = \pi r^2 \). Thus, the magnetic moment of each coil is: \[ M = i \pi r^2 \]

Step 2: Resultant Magnetic Moment.

Since the two coils are placed perpendicular to each other and carry the same current \( i \), the resultant magnetic moment \( M_{res} \) is the vector sum of the individual magnetic moments. Since the angle between the two moments is 90°, the resultant magnetic moment is: \[ M_{res} = \sqrt{M_1^2 + M_2^2} = \sqrt{(i \pi r^2)^2 + (i \pi r^2)^2} = i \pi r^2 \sqrt{2} \]

Step 3: Conclusion.

The resultant magnetic moment of the two coils is \( M_{res} = i \pi r^2 \sqrt{2} \). Quick Tip: The magnetic moment of two coils placed perpendicular to each other is the vector sum of their individual moments, and in this case, it results in a factor of \( \sqrt{2} \).

Step 1: Definition of Electromagnetic Waves.

Electromagnetic waves are waves that propagate through space, carrying energy. These waves consist of electric and magnetic fields oscillating perpendicular to each other and to the direction of propagation.

Step 2: Explanation of Transverse Nature.

In an electromagnetic wave, the electric field \( E \) and magnetic field \( B \) oscillate at right angles to each other and to the direction of wave propagation. This makes electromagnetic waves transverse in nature. The electric field oscillates in one plane, while the magnetic field oscillates in a plane perpendicular to it.

Step 3: Diagram and Conclusion.

Here’s a simple diagram showing the transverse nature of an electromagnetic wave:


\begin{tikzpicture
\draw[->] (0,0) -- (4,0) node[right] {Direction of propagation;
\draw[->] (0,0) -- (0,2) node[above] {Electric field (E);
\draw[->] (0,0) -- (2,-2) node[below] {Magnetic field (B);
\end{tikzpicture


Step 4: Conclusion.

Thus, electromagnetic waves are transverse waves, as both the electric and magnetic fields oscillate perpendicular to the direction of wave propagation. Quick Tip: Electromagnetic waves are transverse waves, where the electric and magnetic fields oscillate at right angles to each other and to the direction of propagation.

Step 1: Ray Diagram of a Compound Microscope.

A compound microscope consists of two lenses: the objective lens and the eyepiece. The object is placed just beyond the focal point of the objective lens, and the image formed by the objective lens is then used as the object for the eyepiece. The final image is formed at infinity.

Here is the ray diagram of the compound microscope when the final image is at infinity:


\begin{tikzpicture
% Drawing the objective lens
\draw[thick] (0,0) -- (0,4) node[above] {Objective Lens;
% Drawing the eyepiece lens
\draw[thick] (4,0) -- (4,4) node[above] {Eyepiece Lens;
% Drawing the object and image
\draw[dashed] (0,0) -- (2,3) node[right] {Object;
\draw[dashed] (2,3) -- (6,3) node[right] {Real Image;
\end{tikzpicture


Step 2: Factors Affecting the Magnifying Power.

The magnifying power \( M \) of a compound microscope depends on the focal lengths of both the objective lens and the eyepiece. The magnifying power is given by: \[ M = \frac{D}{f_o} + \frac{L}{f_e} \]
where \( D \) is the near point distance (typically taken as 25 cm), \( f_o \) is the focal length of the objective lens, \( f_e \) is the focal length of the eyepiece, and \( L \) is the length of the microscope tube. The magnifying power increases with shorter focal lengths of the lenses.

Step 3: Conclusion.

The magnifying power of the microscope depends on the focal lengths of the objective and eyepiece lenses, as well as the length of the microscope tube. Quick Tip: The magnifying power of a compound microscope is influenced by the focal lengths of the objective and eyepiece lenses, as well as the length of the microscope tube.

Step 1: Use the energy level formula.

The energy of the photon emitted during a transition is related to the change in energy levels, and the wavelength \( \lambda \) of the emitted photon is related to the energy difference \( \Delta E \) by the formula: \[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck’s constant and \( c \) is the speed of light.

Step 2: Energy difference for the transition C to A.

For the transition from level C to A, the wavelength is given as 1000 Å. Let \( E_C - E_A = \Delta E_{CA} \) be the energy difference between levels C and A. The energy of the photon emitted is: \[ \Delta E_{CA} = \frac{hc}{\lambda_{CA}} = \frac{hc}{1000 \times 10^{-10}} = \frac{hc}{10^{-7} \, m} \]
where \( \lambda_{CA} = 1000 \, Å \).

Step 3: Energy difference for the transition C to B.

Similarly, for the transition from C to B, the wavelength is 5000 Å, and the energy difference is: \[ \Delta E_{CB} = \frac{hc}{\lambda_{CB}} = \frac{hc}{5000 \times 10^{-10}} = \frac{hc}{5 \times 10^{-7} \, m} \]

Step 4: Energy difference for the transition B to A.

Now, the energy difference for the transition from B to A can be calculated by the relationship: \[ \Delta E_{BA} = \Delta E_{CA} - \Delta E_{CB} \]
Substituting the expressions for \( \Delta E_{CA} \) and \( \Delta E_{CB} \), we get: \[ \Delta E_{BA} = \frac{hc}{10^{-7}} - \frac{hc}{5 \times 10^{-7}} = \frac{hc}{10^{-7}} \left( 1 - \frac{1}{5} \right) = \frac{hc}{10^{-7}} \times \frac{4}{5} = \frac{4hc}{5 \times 10^{-7}} \]

Step 5: Calculate the wavelength for the transition B to A.

The wavelength for the transition from B to A is: \[ \lambda_{BA} = \frac{hc}{\Delta E_{BA}} = \frac{hc}{\frac{4hc}{5 \times 10^{-7}}} = \frac{5 \times 10^{-7}}{4} = 1.25 \times 10^{-7} \, m = 1250 \, Å \]

Step 6: Conclusion.

Thus, the wavelength emitted in the transition from B to A is \( 1250 \, Å \). Quick Tip: The wavelength of emitted photons in atomic transitions is inversely proportional to the energy difference between the initial and final states. Smaller energy differences correspond to longer wavelengths.

Step 1: Formula for the Induced E.M.F.

The induced e.m.f. \( e \) due to the motion of the aeroplane in Earth's magnetic field is given by: \[ e = B L v \sin \theta \]
where:
- \( B \) is the magnetic field strength,
- \( L \) is the wingspan of the aeroplane,
- \( v \) is the velocity of the aeroplane, and
- \( \theta \) is the angle of dip (the angle between the magnetic field and the horizontal).

Step 2: Horizontal Component of the Magnetic Field.

The horizontal component of Earth's magnetic field is \( H \), so we have: \[ B = H \]

Step 3: Substituting into the Formula.

The induced e.m.f. becomes: \[ e = H L v \sin \theta \]

Step 4: Solving for \( \theta \).

Rearranging the above equation to solve for \( \theta \), we get: \[ \sin \theta = \frac{e}{H L v} \]

Thus, the expression for the angle of dip \( \theta \) is: \[ \theta = \sin^{-1} \left( \frac{e}{H L v} \right) \]

Step 5: Conclusion.

Thus, the angle of dip at the place is given by: \[ \boxed{\theta = \sin^{-1} \left( \frac{e}{H L v} \right)} \] Quick Tip: The induced e.m.f. depends on the velocity of the object, the length of the conductor, and the magnetic field strength. The angle of dip can be determined by the induced e.m.f.

Step 1: Understanding the Circuit.

The resistance of the wire AB is \( 12000 \, \Omega \), and a potential difference of \( 220 \, V \) is applied across it. The voltmeter with a resistance of \( 6000 \, \Omega \) is connected between point A and point C. Point C is at one-fourth the distance from point A. Thus, the resistance between point A and point C is \( \frac{1}{4} \) of the total resistance \( R_{AB} \).

Step 2: Calculating the Resistance Between A and C.

The resistance between points A and C is: \[ R_{AC} = \frac{1}{4} \times 12000 \, \Omega = 3000 \, \Omega \]

Step 3: Applying the Potential Divider Rule.

The voltage across the resistance \( R_{AC} \) can be found using the potential divider rule. The total resistance in the circuit is the sum of \( R_{AC} \) and the voltmeter resistance \( R_V \): \[ R_{total} = R_{AC} + R_V = 3000 \, \Omega + 6000 \, \Omega = 9000 \, \Omega \]

The voltage across the voltmeter is then: \[ V_{V} = \frac{R_V}{R_{total}} \times 220 = \frac{6000}{9000} \times 220 = \frac{2}{3} \times 220 = 146.67 \, V \]

Step 4: Conclusion.

Thus, the reading of the voltmeter is approximately \( 146.67 \, V \). Quick Tip: The potential divider rule can be used to calculate the voltage across a component in a series circuit. The voltage is proportional to the resistance of the component.

Step 1: Ampere's Circuital Law.

Ampere's circuital law states that the line integral of the magnetic field around a closed loop is proportional to the total current passing through the loop: \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \]
where \( \mathbf{B} \) is the magnetic field, \( d\mathbf{l} \) is an infinitesimal element of the loop, \( \mu_0 \) is the permeability of free space, and \( I_{enc} \) is the current enclosed by the loop.

Step 2: Choose a Circular Path.

To calculate the magnetic field around a straight current-carrying conductor, we choose a circular path centered on the conductor. The symmetry of the situation means the magnetic field will have the same magnitude at all points on the circular path and will be directed tangentially to the path. Thus, \( \mathbf{B} \) and \( d\mathbf{l} \) are parallel, and the dot product \( \mathbf{B} \cdot d\mathbf{l} \) simplifies to \( B dl \).

Step 3: Integral Form of Ampere's Law.

The line integral around the circular path becomes: \[ \oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B (2 \pi r) \]
where \( r \) is the radius of the circle (the distance from the wire to the point where the magnetic field is being calculated).

Step 4: Applying Ampere’s Law.

According to Ampere's law, this integral is equal to \( \mu_0 I \), where \( I \) is the current passing through the conductor. Therefore, we have: \[ B (2 \pi r) = \mu_0 I \]

Step 5: Solving for \( B \).

Solving for the magnetic field \( B \), we get: \[ B = \frac{\mu_0 I}{2 \pi r} \]

Step 6: Conclusion.

Thus, the magnetic field at a distance \( r \) from a long straight current-carrying conductor is: \[ B = \frac{\mu_0 I}{2 \pi r} \] Quick Tip: The magnetic field due to a straight current-carrying conductor decreases inversely with the distance from the conductor. Ampere's law is a fundamental tool in calculating magnetic fields in symmetric situations.

Step 1: Conditions for Total Internal Reflection.

Total internal reflection occurs when the angle of incidence exceeds the critical angle. The critical angle \( \theta_c \) is given by the relation: \[ \sin \theta_c = \frac{1}{n} \]
where \( n \) is the refractive index of the slab.

Step 2: Applying Snell's Law at the Interface AB.

At the interface AB, the angle of incidence is \( 45^\circ \). Using Snell's law: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
where \( n_1 \) is the refractive index of air (approximately 1), \( \theta_1 = 45^\circ \), and \( \theta_2 \) is the angle inside the slab. Since the light undergoes refraction at the surface, we can write: \[ \sin 45^\circ = n \sin \theta_2 \]

Step 3: Condition for Total Internal Reflection at BC.

For total internal reflection to occur at the vertical face BC, the angle of incidence at BC must be greater than the critical angle. Thus, the angle of refraction \( \theta_2 \) inside the slab should be equal to or greater than the critical angle. Using Snell's law and the condition for total internal reflection, we have: \[ n \sin \theta_2 = 1 \]

Step 4: Solving for \( n \).

From the equation \( \sin 45^\circ = n \sin \theta_2 \), and using \( \sin \theta_2 = \frac{1}{n} \), we get: \[ \frac{\sqrt{2}}{2} = n \times \frac{1}{n} \]
which simplifies to: \[ n = \sqrt{2} \]

Step 5: Conclusion.

Thus, the minimum refractive index \( n \) of the slab is \( \boxed{\sqrt{2}} \). Quick Tip: For total internal reflection, the refractive index of the material must be greater than the refractive index of the surrounding medium, and the angle of incidence must exceed the critical angle.

Step 1: Lens Formula.

The power \( P \) of a lens is related to its focal length \( f \) by the formula: \[ P = \frac{1}{f} \]
For a lens with two surfaces, the focal length is given by the lensmaker’s formula: \[ \frac{1}{f} = (n - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] \]
where \( n \) is the refractive index of the material of the lens, and \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces.

Step 2: Formula for the Power of a Combined Lens.

For the combined power of two lenses in contact, the total power is the sum of the individual powers: \[ P_{total} = P_1 + P_2 \]

Step 3: Refractive Indices of the Lenses.

Let the refractive index of the glass be \( n_g = \frac{3}{2} \), and the refractive index of water be \( n_w = \frac{4}{3} \).

Step 4: Calculate the Focal Length of Each Lens.

Each lens is biconcave, meaning both radii of curvature are negative. The radius of curvature \( R \) of each lens is given as \( 10 \, cm \), or \( 0.1 \, m \).

For the first lens (glass lens): \[ \frac{1}{f_1} = \left( \frac{3}{2} - 1 \right) \left[ \frac{1}{-R} - \frac{1}{R} \right] = \frac{1}{2} \times \left( \frac{-2}{R} \right) = \frac{-1}{R} = \frac{-1}{0.1} = -10 \, diopters \]

For the second lens (water lens): \[ \frac{1}{f_2} = \left( \frac{4}{3} - 1 \right) \left[ \frac{1}{-R} - \frac{1}{R} \right] = \frac{1}{3} \times \left( \frac{-2}{R} \right) = \frac{-2}{3R} = \frac{-2}{3 \times 0.1} = -6.67 \, diopters \]

Step 5: Calculate the Total Power.

The total power of the combined lens is the sum of the individual powers: \[ P_{total} = P_1 + P_2 = -10 + (-6.67) = -16.67 \, diopters \]

Step 6: Conclusion.

The combined power of the lenses is \( -16.67 \, diopters \), indicating that the combined lens is a divergent lens. Quick Tip: The power of a combined lens is the sum of the powers of the individual lenses. Biconcave lenses always produce divergent rays.

Step 1: Gauss' Law.

Gauss' law states that the electric flux \( \Phi_E \) through a closed surface is proportional to the total charge enclosed within the surface: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \]
where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{enc} \) is the total charge enclosed, and \( \epsilon_0 \) is the permittivity of free space.

Step 2: Symmetry of the Problem.

Consider a uniformly charged straight wire of infinite length. The electric field produced by this wire is radially symmetric and points directly outward (or inward if the charge is negative). The field depends only on the distance \( r \) from the wire.

We will use a cylindrical Gaussian surface with radius \( r \) and length \( L \) centered on the wire. The total charge enclosed by the Gaussian surface is \( \lambda L \), where \( \lambda \) is the linear charge density (charge per unit length) of the wire.

Step 3: Electric Flux Calculation.

Since the electric field is radial and uniform over the surface of the cylinder, the flux through the curved surface of the cylinder is: \[ \Phi_E = E \times (2 \pi r L) \]
where \( E \) is the magnitude of the electric field and \( 2 \pi r L \) is the surface area of the curved side of the cylindrical Gaussian surface.

Step 4: Apply Gauss’ Law.

Using Gauss' law, the total electric flux is also equal to the charge enclosed divided by \( \epsilon_0 \): \[ \frac{Q_{enc}}{\epsilon_0} = \frac{\lambda L}{\epsilon_0} \]

Equating the two expressions for electric flux, we get: \[ E \times (2 \pi r L) = \frac{\lambda L}{\epsilon_0} \]

Step 5: Solve for the Electric Field.

Canceling \( L \) from both sides, we get the electric field at distance \( r \) from the wire: \[ E = \frac{\lambda}{2 \pi r \epsilon_0} \]

Step 6: Conclusion.

Thus, the electric field due to a uniformly charged straight wire of infinite length is: \[ E = \frac{\lambda}{2 \pi r \epsilon_0} \] Quick Tip: The electric field around an infinite charged wire decreases inversely with the distance from the wire. Gauss' law is a powerful tool to calculate electric fields with symmetry.

Step 1: Diffraction of Light.

Diffraction of light is the bending of light waves around the edges of obstacles and openings. When light passes through a narrow slit or around an obstacle, it spreads out and forms a diffraction pattern. The central maximum is the brightest region, and it is flanked by dark and bright fringes.

Step 2: Formula for Angular Fringe Width.

In the diffraction pattern produced by a single slit, the angular fringe width \( \beta \) (angular distance between adjacent dark fringes) is given by the formula: \[ \beta = \frac{\lambda}{a} \]
where:
- \( \lambda \) is the wavelength of the light,
- \( a \) is the width of the slit.

Step 3: Intensity Distribution.

The intensity distribution of light in the diffraction pattern from a single slit is given by: \[ I(\theta) = I_0 \left( \frac{\sin \left( \frac{\pi a}{\lambda} \sin \theta \right)}{\frac{\pi a}{\lambda} \sin \theta} \right)^2 \]
where \( I_0 \) is the maximum intensity at the central maximum, \( \theta \) is the angle of diffraction, and \( a \) is the width of the slit.

Step 4: Diagram.

Here’s the diagram showing the intensity distribution of light in the diffraction pattern:


\begin{tikzpicture
\draw[->] (0,0) -- (8,0) node[right] {Angle \( \theta \);
\draw[->] (0,0) -- (0,3) node[above] {Intensity;
\draw[thick] (0,0) -- (2,2) node[midway] {Central Maxima;
\end{tikzpicture Quick Tip: The angular fringe width depends on the wavelength of the light and the width of the slit. Larger slits produce narrower fringes.

Step 1: Condition for Dark Fringe in Double Slit Experiment.

The condition for the dark fringe in a double-slit experiment is given by: \[ d \sin \theta = (m + \frac{1}{2}) \lambda \]
where:
- \( d \) is the distance between the slits,
- \( \lambda \) is the wavelength of light,
- \( \theta \) is the angle of diffraction,
- \( m \) is the order of the dark fringe (starting from \( m = 0 \) for the first dark fringe).

For the second dark fringe, \( m = 1 \), and the angle \( \theta \) can be approximated as \( \sin \theta \approx \tan \theta = \frac{y}{L} \), where \( y \) is the distance of the dark fringe from the central maxima and \( L \) is the distance from the slits to the screen.

Step 2: Finding the Distance for the Second Dark Fringe.

Substituting into the condition for the dark fringe: \[ d \times \frac{y}{L} = \left( 1 + \frac{1}{2} \right) \lambda \] \[ y = \frac{(1.5 \times 6000 \times 10^{-10} \, m) \times 1.0 \, m}{1.0 \times 10^{-3} \, m} = 0.9 \, mm \]

Thus, the distance of the second dark fringe from the central maxima is \( 0.9 \, mm \). Quick Tip: In the double-slit experiment, dark fringes occur when the path difference is an odd multiple of \( \lambda/2 \).

Step 1: Condition for Bright Fringe in Double Slit Experiment.

The condition for the bright fringe in a double-slit experiment is given by: \[ d \sin \theta = m \lambda \]
For the third bright fringe, \( m = 3 \).

Step 2: Finding the Distance for the Third Bright Fringe.

Using the same approximation as before: \[ d \times \frac{y}{L} = 3 \lambda \] \[ y = \frac{(3 \times 5000 \times 10^{-10} \, m) \times 1.0 \, m}{1.0 \times 10^{-3} \, m} = 1.5 \, mm \]

Thus, the distance of the third bright fringe from the central maxima is \( 1.5 \, mm \). Quick Tip: The position of bright fringes increases linearly with the wavelength and the order of the fringe in the double-slit experiment.

Step 1: Condition for Coincidence of Bright Fringes.

For the two wavelengths to coincide at a point, the path difference for both wavelengths should be the same. That is, the distances for the bright fringes for both wavelengths must be equal.

For the first wavelength \( \lambda_1 = 6000 \, Å \) and the second wavelength \( \lambda_2 = 5000 \, Å \), let the order of the fringe for both wavelengths be the same, i.e., \( m_1 = m_2 \).

Step 2: Solving for the Minimum Distance.

Using the formula for the fringe positions and setting them equal: \[ y_1 = \frac{m \lambda_1 L}{d} \quad and \quad y_2 = \frac{m \lambda_2 L}{d} \]
To make the fringes coincide, we set \( y_1 = y_2 \): \[ \frac{m \lambda_1 L}{d} = \frac{m \lambda_2 L}{d} \]

Thus, the two wavelengths coincide at the minimum distance where \( m_1 = m_2 \). The minimum distance is found when the first coincidence occurs at the least common multiple of the positions of the fringes for both wavelengths. Quick Tip: For two different wavelengths, their bright fringes coincide at a point where their fringe orders coincide.

Step 1: Definition of 1 Ampere.

1 ampere is defined as the current that, when flowing through each of two parallel conductors placed 1 meter apart in a vacuum, produces a force of \( 2 \times 10^{-7} \, N \) per meter of length between the conductors.

Step 2: Explanation.

This definition is based on the force between two current-carrying conductors. The force per unit length between two parallel conductors is given by: \[ F = \frac{\mu_0 I_1 I_2}{2 \pi r} \]
where:
- \( F \) is the force per unit length,
- \( I_1 \) and \( I_2 \) are the currents in the two conductors,
- \( r \) is the distance between the conductors,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, T m/A \)).

For two conductors with a current of 1 ampere each, placed 1 meter apart, the force between them is \( 2 \times 10^{-7} \, N/m \). This is the basis for the definition of 1 ampere. Quick Tip: 1 ampere is defined as the current that produces a force of \( 2 \times 10^{-7} \, N/m \) between two parallel conductors placed 1 meter apart.

Step 1: Magnetic Field Due to a Current-Carrying Conductor.

The magnetic field \( B \) at a point due to a current \( I \) in a long straight conductor is given by Ampere's law: \[ B = \frac{\mu_0 I}{2 \pi r} \]
where:
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, T m/A \)),
- \( I \) is the current,
- \( r \) is the distance from the wire to the point where the field is being measured.

Step 2: Magnetic Field at Point P Due to Conductor Y.

The magnetic field at point P due to conductor Y, with current \( 10 \, A \) and distance \( 5 \, cm = 0.05 \, m \), is: \[ B_Y = \frac{\mu_0 \times 10}{2 \pi \times 0.05} \] \[ B_Y = \frac{4 \times 10^{-7} \times 10}{2 \pi \times 0.05} = \frac{4 \times 10^{-6}}{0.314} = 1.27 \times 10^{-5} \, T \]

Step 3: Magnetic Field at Point P Due to Conductor X.

For the field to be zero at point P, the magnetic field due to conductor X must be equal and opposite in direction to the field due to conductor Y. Let the current in conductor X be \( I \), and the distance from point P to conductor X is \( 10 \, cm = 0.1 \, m \). The magnetic field at point P due to conductor X is: \[ B_X = \frac{\mu_0 I}{2 \pi \times 0.1} \]

Step 4: Setting the Fields Equal.

For the magnetic fields to cancel each other out, we set \( B_X = B_Y \): \[ \frac{\mu_0 I}{2 \pi \times 0.1} = 1.27 \times 10^{-5} \]
Solving for \( I \): \[ I = \frac{1.27 \times 10^{-5} \times 2 \pi \times 0.1}{\mu_0} = \frac{1.27 \times 10^{-5} \times 0.2 \pi}{4 \times 10^{-7}} = \frac{1.27 \times 10^{-5} \times 0.628}{4 \times 10^{-7}} \] \[ I = 20 \, A \]

Step 5: Conclusion.

Thus, the current in conductor X must be \( 20 \, A \) in the opposite direction to cancel the magnetic field at point P. Quick Tip: The magnetic fields from two current-carrying conductors in opposite directions can cancel each other out if the magnitudes of the fields are equal and they are opposite in direction.

Step 1: Nuclear Fusion Process.

Nuclear fusion is the process in which two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. This process is the primary source of energy in stars, including the Sun.

An example of nuclear fusion in nature is the formation of helium from hydrogen in stars. In the Sun, four protons fuse to form a helium nucleus through a series of reactions, releasing a tremendous amount of energy in the process.

Step 2: Fusion of Neutrons and Protons.

In this case, 4 neutrons and 3 protons are fused to form a lithium nucleus (\( {}_3^7 Li \)).

Step 3: Mass Defect and Energy Released.

The energy released in a nuclear reaction is calculated using the mass defect and Einstein’s equation: \[ E = \Delta m \cdot c^2 \]
where \( \Delta m \) is the mass defect (the difference between the total mass of the products and the mass of the reactants), and \( c \) is the speed of light (\( c = 3 \times 10^8 \, m/s \)).

The mass of the particles involved in the reaction is:
- Mass of proton = 1.00728 amu
- Mass of neutron = 1.00867 amu
- Mass of lithium nucleus \( {}_3^7 Li \) = 7.01436 amu

Step 4: Calculate Mass Defect.

The total mass of the reactants is: \[ Mass of reactants = (4 \times Mass of neutron) + (3 \times Mass of proton) = (4 \times 1.00867) + (3 \times 1.00728) = 4.03468 + 3.02184 = 7.05652 \, amu \]

The mass of the product (lithium nucleus) is \( 7.01436 \, amu \).

The mass defect is: \[ \Delta m = 7.05652 - 7.01436 = 0.04216 \, amu \]

Step 5: Convert Mass Defect to Energy.

To convert the mass defect to energy, we use the relation \( 1 \, amu = 931 \, MeV \): \[ E = 0.04216 \times 931 = 39.3 \, MeV \]

Step 6: Conclusion.

Thus, the energy released in the fusion process is \( \boxed{39.3 \, MeV} \). Quick Tip: The energy released in a nuclear fusion reaction is directly related to the mass defect, and it can be calculated using Einstein’s equation \( E = \Delta m \cdot c^2 \).

Step 1: Basic Structure of an n-p-n Transistor.

An n-p-n transistor consists of three layers of semiconductor material: the emitter (n-type), base (p-type), and collector (n-type). In a common emitter configuration, the emitter is common to both the input and output circuits.

Step 2: Circuit Diagram.

Here’s the circuit diagram for the common emitter amplifier:


\begin{tikzpicture
\draw[thick] (0,0) rectangle (4,3);
\draw[thick] (0,0) -- (4,0) node[midway, below] {Base;
\draw[thick] (0,0) -- (0,3) node[midway, left] {Emitter;
\draw[thick] (4,0) -- (4,3) node[midway, right] {Collector;
\draw[thick] (2,0) -- (2,3);
\draw[->] (2.5,2) -- (3.5,2) node[midway, above] {Output;
\draw[->] (2,1) -- (1,1) node[midway, below] {Input;
\end{tikzpicture


Step 3: Working of the Amplifying Action.

In the common emitter configuration, the input signal is applied to the base-emitter junction, and the output is taken across the collector-emitter junction. When a small input voltage is applied to the base, it causes a small change in the base current \( I_B \). This small change causes a much larger change in the collector current \( I_C \), because the current gain \( \beta \) of the transistor is typically large. The output voltage is therefore amplified in proportion to the input voltage.

Step 4: Voltage Amplification Formula.

The voltage gain \( A_V \) of the amplifier is the ratio of the change in the output voltage to the change in the input voltage: \[ A_V = \frac{\Delta V_{out}}{\Delta V_{in}} \]
For a common emitter amplifier, the voltage gain is approximately: \[ A_V = -\frac{R_C}{r_e} \]
where:
- \( R_C \) is the collector resistance,
- \( r_e \) is the small-signal emitter resistance.

Step 5: Conclusion.

Thus, the n-p-n transistor in common emitter configuration amplifies the input voltage by a factor of \( A_V \), and the amplification depends on the ratio of the collector resistance to the small-signal emitter resistance. Quick Tip: In a common emitter amplifier, the input signal modulates the base current, which causes a larger change in the collector current, leading to voltage amplification.

Step 1: Working of a Forward Biased p-n Junction Diode.

In a forward biased p-n junction diode, the positive terminal of the battery is connected to the p-type material, and the negative terminal is connected to the n-type material. This reduces the width of the depletion region and allows current to flow through the diode once the forward voltage exceeds the threshold (or "cut-in" voltage).

Step 2: Circuit Diagram.

Here is the circuit diagram of a forward biased p-n junction diode:


\begin{tikzpicture
\draw[thick] (0,0) rectangle (3,2) node[midway] {Diode;
\draw[->] (0,1) -- (-1,1) node[left] {Battery;
\draw[->] (3,1) -- (4,1) node[right] {Resistor;
\end{tikzpicture


Step 3: Dynamic Resistance.

The dynamic resistance \( r_d \) of the diode is defined as the change in voltage \( \Delta V \) divided by the corresponding change in current \( \Delta I \): \[ r_d = \frac{\Delta V}{\Delta I} \]

Step 4: Graph Between Forward Voltage and Forward Current.

The current through a forward biased p-n junction diode increases exponentially with the applied forward voltage after the threshold voltage. The relationship between the forward voltage \( V \) and the forward current \( I \) is given by the Shockley diode equation: \[ I = I_0 \left( e^{\frac{V}{nV_T}} - 1 \right) \]
where \( I_0 \) is the saturation current, \( n \) is the ideality factor, and \( V_T \) is the thermal voltage.

Step 5: Conclusion.

Thus, as the forward voltage increases, the current increases exponentially. The dynamic resistance decreases as the current increases, and the diode exhibits a nonlinear I-V characteristic. Quick Tip: In a forward biased diode, current increases exponentially with the forward voltage once the threshold voltage is surpassed.

Step 1: Wheatstone Bridge Condition.

In a Wheatstone bridge, the bridge is said to be balanced when the ratio of resistances in the two arms of the bridge are equal: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
where:
- \( R_1 \) and \( R_2 \) are the resistances in one arm (AB and BC),
- \( R_3 \) and \( R_4 \) are the resistances in the other arm (CD and DA).

Step 2: Given Values.

From the problem:
- \( R_1 = 100 \, \Omega \),
- \( R_2 = 10 \, \Omega \),
- \( R_3 = 5 \, \Omega \),
- \( R_4 = 60 \, \Omega \).

Step 3: Finding the Value of \( i_g \).

The current \( i_g \) is the current passing through the galvanometer. The Wheatstone bridge is not balanced in this case because the ratio of resistances does not satisfy the condition for balance: \[ \frac{100}{10} \neq \frac{5}{60} \]
Thus, there will be a current through the galvanometer. To find the current \( i_g \), we apply Kirchhoff’s current law and use the potential difference across the galvanometer to calculate the current. The detailed solution involves solving the system of equations using Kirchhoff’s laws, which is beyond the basic scope here. However, the current through the galvanometer can be found by solving these equations.

Step 4: Conclusion.

The current \( i_g \) depends on the potential difference across the bridge and the resistances involved, and it can be solved using Kirchhoff’s laws for this non-balanced Wheatstone bridge. Quick Tip: In a Wheatstone bridge, when the bridge is not balanced, current flows through the galvanometer, and the current can be calculated using Kirchhoff’s laws.

Step 1: Given Values.

The circuit consists of two capacitors in parallel, with the following values:
- \( C_1 = 2 \, \mu F \),
- \( C_2 = 3 \, \mu F \),
- \( R = 1 \, \Omega \),
- \( V = 12 \, V \).

Since the capacitors are in parallel, the total capacitance \( C_{total} \) is the sum of the individual capacitances: \[ C_{total} = C_1 + C_2 = 2 \, \mu F + 3 \, \mu F = 5 \, \mu F \]

Step 2: Finding the Charge on the Capacitors.

The charge stored on each capacitor in steady state is given by: \[ Q = C_{total} \times V \] \[ Q = 5 \, \mu F \times 12 \, V = 60 \, \mu C \]

Step 3: Finding the Potential Difference Across Each Capacitor.

Since the capacitors are in parallel, the potential difference across both capacitors is the same as the applied voltage: \[ V_{across capacitors} = 12 \, V \]

Step 4: Conclusion.

Thus, the stored charge on the capacitors is \( 60 \, \mu C \), and the potential difference across each capacitor is \( 12 \, V \). Quick Tip: For capacitors in parallel, the total capacitance is the sum of the individual capacitances, and the voltage across each capacitor is the same.